It is proved that the monotonicity of the symmetric interval of even function is opposite Ah You know, you know, There are only two things I don't understand. I'm in a hurry Let y = f (x) be an even function, then f (- x) = f (x) If X1 > x2 > 0, then - x1f (x2), that is, f (x) decreases In the same way, it can be proved that f (x) is a decreasing function on the positive half axis and an increasing function on the negative half axis In this case, why is it necessary to assume that the positive semiaxis is monotonically increasing? Finally, why is f (- x1) > F (x2), that is, f (x) decreasing? How to judge?

It is proved that the monotonicity of the symmetric interval of even function is opposite Ah You know, you know, There are only two things I don't understand. I'm in a hurry Let y = f (x) be an even function, then f (- x) = f (x) If X1 > x2 > 0, then - x1f (x2), that is, f (x) decreases In the same way, it can be proved that f (x) is a decreasing function on the positive half axis and an increasing function on the negative half axis In this case, why is it necessary to assume that the positive semiaxis is monotonically increasing? Finally, why is f (- x1) > F (x2), that is, f (x) decreasing? How to judge?

Let y = f (x) be an even function, then f (- x) = f (x)
If X1 > x2 > 0, then - x1f (x2), that is, f (x) decreases
In the same way, it can be proved that f (x) is a decreasing function on the positive half axis and an increasing function on the negative half axis
In this case, why is it necessary to assume that the positive semiaxis is monotonically increasing?
Answer: "may as well set" means that it can be set as monotonic increasing or monotonic decreasing
Let f (x) monotonically decrease on the positive semiaxis of X, then f (x 1) < f (x 2),
So f (- x1) < f (- x2), that is, f (x) increases in the negative half axis of X
You can set f (x) as a simple function, such as: X & sup2;, - X & sup2; these two functions increase monotonically on the positive half axis and decrease monotonically on the negative half axis
Finally, why is f (- x1) > F (x2), that is, f (x) decreasing?
There is a minus sign f (- x1) > F (- x2) missing here, where - X1 and - x2 are on the negative half axis of X, that is
-x1